16
votes
Accepted
The Lévy-Khintchine formula and integrability conditions of a random measure
The condition
$$\int_{\mathbb{R} \backslash \{0\}} \min\{1,z^2\} \nu(dz)<\infty$$
is equivalent to
$$\int_{|z| \leq 1} z^2 \, \nu(dz) < \infty \quad \text{and} \quad \int_{|z| \geq 1} \nu(dz) ...
- 113k
11
votes
Accepted
Proving properties for the Poisson-process.
Let $(X_t)_{t \geq 0}$ be a Poisson process with intensity $\lambda$.
Step 1: $(X_t)_{t \geq 0}$ has almost surely increasing sample paths.
Proof: Fix $s \leq t$. Since $(X_t)_{t \geq 0}$ has ...
- 113k
10
votes
Accepted
Jumps of Lévy process
Yes, there is a very strong relation between the (distribution of the) jumps of a Lévy process and its Lévy measure. In fact, the Lévy measure describes the jump behaviour of the corresponding Lévy ...
- 113k
7
votes
Good book that contains stochastic integration, martingales and Lévy-processes?
Since Lévy processes are used a lot in finance, there are several books on this topic (that is, Lévy processes and their applications in finance). For example "Stochastic Calculus for Finance II" by S....
- 113k
7
votes
Accepted
Etemadi's inequality
The inequality which both ziT and I used is a direct consequence of the following inequality.
Lemma 1: Let $X_1,\ldots,X_n$ be independent random variables and $S_k := \sum_{j=1}^k X_j$, $k=1,\...
- 113k
6
votes
Application of Lévy–Khinchine formula
Depending on what you know about the Lévy triplet it is much easier to calculate the characteristic functions using the very definition of Wiener Process and Poisson process, respectively. For example,...
- 113k
5
votes
Accepted
Stochastic Continuity of a Lévy Process
By the homogeneity of the increments, we have
$$\mathbb{P}(|X(s)-X(t)|>\varepsilon) = \mathbb{P}(|X(t-s)|>\varepsilon).$$
This means that
$$\lim_{s \to t} \mathbb{P}(|X(s)-X(t)|>\...
- 113k
5
votes
Accepted
Largest jumps of a spectrally positive $\alpha$-stable process
Denote by $N$ the jump measure of the Lévy process, i.e. $$N_t(B) := N([0,t] \times B) := \sharp \{s \in [0,t]; \Delta X_s := X_s-X_{s-} \in B\},$$ and by $\nu$ its Lévy measure. It is widely known ...
- 113k
5
votes
Accepted
Central Limit Theorem for Lévy Process
Without any additional assumptions on the Lévy process $(X_t)_{t \geq 0}$, a central limit theorem does not hold true.
Let $(X_t)_{t \geq 0}$ be a (one-dimensional) Lévy process with Lévy triplet $(...
- 113k
5
votes
Accepted
Gaussian process - Hölder continous paths
The Kolmogorov-Chentsov theorem states that if for any $T\gt 0$, there exists $\alpha,\beta$ and $C\gt 0$ such that for any $s,t\in [0,T ]$,
$$ \tag{1} \mathbb E \left |X_t-X_s\right|^ \alpha \...
- 159k
4
votes
Accepted
Finiteness of the number of big jumps of a Lévy process on a finite interval
Let $(Y_t)_{t \in [0,1]}$ be a Lévy process and denote by $\Delta Y_t := Y_t-Y_{t-}$ the jump height at time $t$. The aim is to show that $(Y_t)_{t \in [0,1]}$ has only finitely many jumps with jump ...
- 113k
4
votes
Accepted
Lévy-Khintchine formula and Taylor expansion
The condition ensures that the integral
$$\int_{\mathbb{R} \backslash \{0\}} (e^{iux}-1-iux 1_{\{|x|<1\}}) \, \nu(dx)$$
is well-defined.
Since $|e^{iux}| \leq 1$, we have
$$|e^{iux}-1| \leq 2$$...
- 113k
4
votes
Left-continuity of a Lévy filtration
Is essentially $\mathcal{F}_t = \mathcal{F}_{t-}+$ null sets?
Yes. Roughly speaking, $\mathcal{F}_t$ contains all the information in $\mathcal{F}_{t-}$ plus the information for which $\omega$ a jump ...
- 113k
4
votes
Good book that contains stochastic integration, martingales and Lévy-processes?
It depends a little bit on your interests, but as you might know, stochastic processes and Itô-calculus is excessively used in quantitative finance. I can recommend some books which really explain the ...
- 1,350
4
votes
Accepted
Integration with respect to a Poisson random measure
(1) $\iff$ (2)
Since $f \geq 0$ is measurable, there exists a sequence $(f_n)_{n \in \mathbb{N}}$ of simple functions (aka elemtary functions/step functions) such that $0 \leq f_n \uparrow f$. Each $...
- 113k
4
votes
Accepted
Why is $T_n-T_{n-1}$ independent of $\mathcal{F}_{T_{n-1}}$ where $T_{n}=\inf \{t>T_{n-1}: |X_t-X_{T_{n-1}}| \geq C\}$?
Let $T$ be a stopping time and $Y_t := X_{T+t}-X_T$ the restarted Lévy process. Theorem 32 states that the $\sigma$-algebra generated by $(Y_t)_{t \geq 0}$ is independent of $\mathcal{F}_T$, i.e.
$$\...
- 113k
4
votes
Accepted
What is the true definition of a Lévy process?
Typically one distinguishes between "Lévy processes" and "Lévy processes in law".
Definition: Let $(X_t)_{t \geq 0}$ be a stochastic process such that $X_0 = 0$ almost surely and $(X_t)_{t \geq 0}$...
- 113k
4
votes
Accepted
Why are processes with stationary independent increments nonstationary?
If $(N_t)_{t \geq 0}$ is a Lévy process (i.e. a stochastic process with stationary and independent increments) and $h \neq 0$, then the random variables $N_t$ and $N_{t+h}$ cannot have the same ...
- 113k
3
votes
Accepted
Change from stochastic exponential to exponential of Lévy process - Applebaum
$\nu \circ f^{-1}$ denotes the push-forward measure (or image measure) of $f$ with respect to $\nu$, i.e.
$$\nu_1(B) = (\nu \circ f^{-1})(B) = \int 1_B(f(x)) \, \nu(dx)$$
for any Borel set $B$. This ...
- 113k
3
votes
Accepted
Quadratic Variation of a square-integrable Lévy process
Hints:
Since $(X_t)_{t \geq 0}$ is a martingale and $X_0=0$, we have $\mathbb{E}(X_t)=0$ for all $t \geq 0$.
If a random variable $Y$ has finite second moment, then $$\mathbb{E}(Y) = \frac{1}{\imath} ...
- 113k
3
votes
Accepted
Intuition behind a set of a assumptions about Lévy processes.
This assumptions means in particular that the jump measure $\nu$ of the Lévy process $(X_t)_{t \geq 0}$ has a density $w$ with respect to the Lebesgue measure. Moreover, the form of the integral shows ...
- 113k
3
votes
Accepted
Blumenthal-Getoor index for IG and $\Gamma$
Example 1 (Gamma process): As @Goulifet pointed out, the Lévy measure of a Gamma process is given by $$\nu(dx) =c x^{-1} e^{-\alpha x} \, dx =: p(x) \, dx.$$ Here $\alpha$ and $c$ are fixed constants. ...
- 113k
3
votes
Accepted
$E[1_{\lbrace P_T-P_{\tau_n}=0\rbrace}\int_{\tau_n}^T h(s)dN_s]=0?$
The answer to the first question is yes :
$X_t=\int_{t}^T h(s)dN_s$ this process in not adapted to the natural filtration of $N$ (as it looks into the future) so we can not apply optiaml sampling ...
- 5,417
3
votes
Random process with stationary independent increments determined by first order distribution?
Note that the equality
$$X_t = 2 X_{t/2}$$
does not hold. It follows from stationarity of the increments that
$$X_{t}-X_{t/2} \stackrel{d}{\sim} X_{t/2},$$
i.e. they are equal in distribution and ...
- 113k
3
votes
Accepted
Subordination of a Levy process when the "subordinator" is not nondecreasing
Usually, Lévy processes $(X_t)_{t \geq 0}$ are considered, i.e. the time $t$ is non-negative. This means that we need the subordinator $(Z_t)_{t \geq 0}$ to satisfy $Z_t \geq 0$. Since Lévy processes ...
- 113k
3
votes
Accepted
Lévy Process existence of the expectation of the supremum of the past process.
Yeah, applying Etemadi's inequality is a good idea. The following identity, which holds for any non-negative random variable $X$, will also be useful:
$$\mathbb{E}(X) = \int_0^{\infty} \mathbb{P}(X &...
- 113k
3
votes
Stochastic Integral with respect to Compensated Poisson Process
This might be helpful: Theorem 29 on page 173 of Protter's "Stochastic integration and differential equations":
Let $M$ be a local martingale and let $H\in \mathcal P$ be locally bounded. then the ...
- 1,088
3
votes
Accepted
Lévy-process property
Yes, this results holds for any Lévy process which is integrable.
It follows from the independence and stationarity of the increments that
$$\mathbb{E}X_{k/n} = \mathbb{E} \left( \sum_{j=1}^k (X_{j/...
- 113k
3
votes
Why is a continuous Lévy process twice integrable?
Let me try to indicate an answer to what I think is your real question ("Is there a direct way to show that a continuous Lévy process is square integrable?") A detailed discussion of this and much ...
- 22.8k
3
votes
Accepted
Ornstein-Uhlenbeck Process simulation bug
If the time vector is t = 0:dt:500; there is an overflow in exp(th*t(i+1)) in the for loop, as t gets large.
Perhaps something like this:
...
- 1,312
Only top scored, non community-wiki answers of a minimum length are eligible